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A wave travelling along uniform string represented by $Y=A \sin (\omega t-k x)$ is superimposed on another wave travelling along the same string represented by $Y=A \sin (\omega t+k x)$. The resultant is
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The correct answer is:
A standing wave having nodes at $x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}$, where $n=0,1,2,3 \ldots \ldots$
The equation of the standing wave formed due to superposition of $Y=A \sin (\omega t-k x)$ and $Y=A \sin (\omega t+k x)$ is given by,
$Y=[2 A \cos (k x)] \sin (\omega t)$
Location of the nodes is dictated by condition:
$[2 A \cos (k x)]=0$
Therefore,
$\begin{aligned} & k x=\left(\frac{2 \pi}{\lambda}\right) x=(2 n+1) \frac{\pi}{2} \\ & \Rightarrow x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2} ; n=0,1,2, \ldots .\end{aligned}$
$Y=[2 A \cos (k x)] \sin (\omega t)$
Location of the nodes is dictated by condition:
$[2 A \cos (k x)]=0$
Therefore,
$\begin{aligned} & k x=\left(\frac{2 \pi}{\lambda}\right) x=(2 n+1) \frac{\pi}{2} \\ & \Rightarrow x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2} ; n=0,1,2, \ldots .\end{aligned}$
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